I am trying to integrate this function with respect to z.
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{z^2}{2}(1-2s)+tz}dz$$ The answer book is telling me it should be:
$$\sqrt{\frac{2 \pi}{1-2s}}$$
If someone has any idea how that's possible that would be awesome! Thanks.
On
You can refer to standard tables of gaussian integrals to get the answer. Typically when solving these types of problems you say "refer to the table of integrals for this gaussian integral". The identity you need can be found on wikipedia: \begin{equation} \int_{-\infty}^{\infty}e^{-ax^2+bx+c}dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}+c} \end{equation}
So for your integral: \begin{equation} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{(1-2s)}{2}z^2+tz}dx \end{equation}
you have $a=\frac{1-2s}{2}$, $b=t$ and $c=0$.
As such your solution is: \begin{equation} \sqrt{\frac{1}{1-2s}}e^{\frac{t^2}{2(1-2s)}} \end{equation}
This has the same form as the answer you provide, however, differs by a multiplicative factor. Either the book is missing this factor or there could be a mistake earlier in your math.
Note that the integral is of the form:
$$\int_{-\infty}^{\infty}e^{-a\phi^2}d\phi=\sqrt{\frac{\pi}{a}}$$
And if you substitute again with $$\phi=z-\frac{t}{1-2s}$$ You'll get the integral into the required form.